Question

Find the coordinates of the centres and the radii of the circles whose equation is:
$\sqrt{1+m^2}(x^2+y^2)-2cx-2mcy=0$.

Answer 1

For a general equation of a circle $x^2+y^2+2gx+2fy+d=0$, the centre is given as $(-g,-f)$

The given equation of circle $\sqrt{1+m^2}(x^2+y^2)-2cx-2mcy=0.$ can be written as

$x^2+y^2-\frac{2c}{\sqrt{1+m^2}}x-\frac{2mc}{\sqrt{1+m^2}}y=0$

Here, $g=\frac{-c}{\sqrt{1+m^2}}$ and

$f=\frac{-mc}{\sqrt{1+m^2}}$, $d=0$

So, coordinates of centre are

$(\frac{c}{\sqrt{1+m^2}},\frac{mc}{\sqrt{1+m^2}})$

And radius is given by

$r=\sqrt{g^2+f^2-d}$

$r=\sqrt{\frac{c^2+m^2{c}^2}{1+m^2}}$

$r=c$